MATLAB在经济学分析中的应用指南

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"MATLAB的经济学分析PDF" 是一本关于使用MATLAB进行经济学分析的教程，旨在帮助读者理解和应用MATLAB解决经济问题。书中涵盖了数学基础、动态响应等多个主题。在数学预习部分，作者P.C.Chau介绍了微分方程模型的基础知识，包括简单微分方程、拉普拉斯变换及其在控制问题中的应用。拉普拉斯变换是信号处理和控制系统理论中的关键工具，它将微分方程转换为代数方程，便于分析系统的稳态和暂态行为。书中详细讨论了初始值和最终值定理，这对于理解系统的动态特性至关重要。此外，还讲解了部分分式展开，这是求解复根或重复根时解拉普拉斯方程的方法。这部分内容分为三个案例，分别对应实数根、复数根和重复实数根的情况。接着，作者介绍了传递函数、极点和零点的概念，这些都是系统分析中的核心概念，它们决定了系统的稳定性、响应速度和振荡特性。书中还总结了极点的特征，并通过两个瞬态模型示例（如一个暂态响应示例和一个搅拌罐加热器）来具体阐述这些概念。线性化非线性方程的方法也在这一章节中被提及，这对于处理实际经济模型中的非线性关系非常有用。最后，块图简化技术是控制系统设计中的重要步骤，可以帮助简化复杂系统模型的表示。在动态响应章节，首先讨论了一阶微分方程模型，包括阶跃响应和脉冲响应，这些都是系统在外部输入变化下的典型行为。一阶模型常用于描述具有积分特性的过程。然后，第二章深入探讨了二阶微分方程模型，特别是其在时间域内的步进响应特性，包括欠阻尼系统的特征。接着，处理带有延迟（死时间）的过程，这是许多经济系统中常见的现象，因为决策和执行之间通常存在时间滞后。对于更高阶的过程，书中有简化的近似方法，以方便处理更复杂的系统。这本书提供了一个全面的MATLAB经济学分析框架，从基本数学原理到实际的经济模型分析，适合希望利用MATLAB进行经济分析的学生和专业人士学习。通过学习书中的内容，读者可以掌握使用MATLAB建模、模拟和分析经济系统的基本技能。

2 - 8

5. Impulse function (Fig. 2.3)

L[δ(t)] = 1, and L[Aδ(t)] = A (2-21)

The (unit) impulse function is called the Dirac (or simply delta) function in mathematics.

If we

suddenly dump a bucket of water into a bigger tank, the impulse function is how we describe the

action mathematically. We can consider the impulse function as the unit rectangular function in

Eq. (2-20) as T shrinks to zero while the height 1/T goes to infinity:

δ(t) = lim

T → 0

u(t) – u(t – T)

The area of this "squeezed rectangle" nevertheless remains at unity:

lim

T → 0

)=1

, or in other words

δ(t) dt=1

–

∞

The impulse function is rarely defined in the conventional sense, but rather via its important

property in an integral:

f(t) δ(t) dt = f(0)

–

∞

, and

f(t)

(t – t

) dt=f(t

)

– ∞

∞

(2-22)

The Laplace transform of the impulse function is obtained easily by taking the limit of the unit

rectangular function transform (2-20) with the use of L'Hospital's rule:

δ(t = lim

→

1 – e

– sT

= lim

→

– sT

From this result, it is obvious that L[Aδ(t)] = A.

2.4 Initial and final value theorems

We now present two theorems which can be used to find the values of the time-domain function at

two extremes, t = 0 and t = ∞, without having to do the inverse transform. In control, we use the

final value theorem quite often. The initial value theorem is less useful. As we have seen from our

very first example in Section 2.1, the problems that we solve are defined to have exclusively zero

initial conditions.

Initial Value Theorem:

lim

s–> ∞

[sF(s)] = lim

t–>0

f(t)

(2-23)

Final Value Theorem:

lim

s–>0

[sF(s)] = lim

t–> ∞

f(t)

(2-24)

The final value theorem is valid provided that a final value exists. The proofs of these theorems are

straightforward. We will do the one for the final value theorem. The proof of the initial value

theorem is in the Review Problems.

Consider the definition of the Laplace transform of a derivative. If we take the limit as s

approaches zero, we find

In mathematics, the unit rectangular function is defined with a height of 1/2T and a width of 2T

from –T to T. We simply begin at t = 0 in control problems. Furthermore, the impulse function is

the time derivative of the unit step function.

2 - 10

2.5 Partial fraction expansion

Since we rely on a look-up table to do reverse Laplace transform, we need the skill to reduce a

complex function down to simpler parts that match our table. In theory, we should be able to

"break up" a ratio of two polynomials in s into simpler partial fractions. If the polynomial in the

denominator, p(s), is of an order higher than the numerator, q(s), we can derive

(s) =

q(s)

p(s)

(s + a

)

(s + a

)

+ ...

(s + a

)

+ ...

(s + a

)

(2-25)

where the order of p(s) is n, and the a

are the negative values of the roots of the equation p(s) = 0.

We then perform the inverse transform term by term:

f(t) =

–1

[F(s)] =

–1

(s + a

)

–1

(s + a

)

+ ...

–1

(s + a

)

+ ...

–1

(s + a

)

(2-26)

This approach works because of the linear property of Laplace transform.

The next question is how to find the partial fractions in Eq. (2-25). One of the techniques is the

so-called Heaviside expansion, a fairly straightforward algebraic method. We will illustrate

three important cases with respect to the roots of the polynomial in the denominator: (1) distinct

real roots, (2) complex conjugate roots, and (3) multiple (or repeated) roots. In a given problem,

we can have a combination of any of the above. Yes, we need to know how to do them all.

✑ 2.5.1 Case 1: p(s) has distinct, real roots

✎ Example 2.4: Find f(t) of the Laplace transform

(s) =

– 12

– 4s – 4)

From Example 2.3, the polynomial in the denominator has roots –1, –2, and +2, values that will

be referred to as poles later. We should be able to write F(s) as

– 12

(s + 1) (s + 2) (s – 2)

(s + 1)

(s + 2)

(s – 2)

The Heaviside expansion takes the following idea. Say if we multiply both sides by (s + 1), we

obtain

– 12

(s + 2) (s – 2)

(s + 2)

(s + 1) +

(s – 2)

(s + 1)

which should be satisfied by any value of s. Now if we choose s = –1, we should obtain

– 12

(s + 2) (s – 2)

s=–1

Similarly, we can multiply the original fraction by (s + 2) and (s – 2), respectively, to find

– 12

(s + 1) (s – 2)

s=–2

and

If the order of q(s) is higher, we need first carry out "long division" until we are left with a

partial fraction "residue." Thus the coefficients α

are also called residues. We then expand this

partial fraction. We would encounter such a situation only in a mathematical problem. The models

of real physical processes lead to problems with a higher order denominator.

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